Suppose an aquifer has become contaminated with radioactive uranium (this has ha
ID: 1003834 • Letter: S
Question
Suppose an aquifer has become contaminated with radioactive uranium (this has happened at some national labs where uranium was being enriched for nuclear weapons, but also in California's Central Valley as an indirect result of nitrate runoff from agriculture that mobilized naturally occurring uranium). The uranium under these conditions is dominated by the soluble uranyl polyatomic cation (UO_z^2+). The Maximum Contamination Limit (MCL) for uranium in drinking water is 30 mu g/L...or in this case 126 nM...dissolved U. A community wants to drill a new drinking water well in this aquifer, and it is your job to maintain this water below the MCL Given the sorption information below, you decide to design and implement an adsorption filtration system at the wellhead to remove the uranium (like a giant Brita filter). The concentration of UO_2^2+ in the aquifer is 2 mu M. The pH of the water in the aquifer is 8. The filtration material has sorption sites (S) at a density of 3 moles of sites per kilogram of sorbent (based on data you find in the literature). The community, about the size of Santa Cruz, needs 40 million liters of water per day. The plan is to change the sorbent once a month, and then dispose of it as hazardous (low-level radioactive) waste. Assuming the uranium reaches equilibrium with the sorbent, how much sorbent must the system contain if you want the water to be right at the MCL after 30 days have elapsed? Ignore activity effects and UO_2^2+ side reactions (in reality, UO_2^2+ forms strong carbonate complexes...making it conservative in seawater). Is this a workable approach? What would you change about the sorbent to improve this design (i.e., the K_a or the K_ads)?Explanation / Answer
Let UO2+2 = U, sorbent = S.
Amount of water required per month = 30x40x1000000 Litres = a
MCL of U = 126 n moles per litre = 126x10-9 moles per litre = b
MCL of U for water for 30 days = a x b = 151.2 moles =d
Conc. Of U per litre in aquifier = 2x10-6 moles = e
Total moles of U for 30 days of water = d x e = 2400 moles.
After leaving U for permissible limit = 2400 – 151.2 = 2248.8 moles of U must absorb.
3 moles of U can be absorbed by 1 kg of S
Mass of S required to absorb 2248.8 moles = 749.6 kg
b) 750 kg of sorbent is practically difficult.
From log K values, it is clear that the dissociation of S-H is weak and formation of S-U is good enough.
Thus, pH = 8 should maintain during the adsorption and sorbent must regenerate in acidic condition.
Precisely, 1 kg of sorbent contains 3 moles of reactive sites.
Since log Ka = -2.8, the [s-] = [H+] = 0.04 moles for each kg of S
(formula: degree of dissociation= sqrt(Ka/C).here ka is given and C = 3 moles per litre)
pH= - log(0.04) = 1.4
sorbent can be generated at pH = 1.4 and can be reused.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.